       Re: UnitStep

• To: mathgroup at smc.vnet.net
• Subject: [mg52377] Re: UnitStep
• From: "David W. Cantrell" <DWCantrell at sigmaxi.org>
• Date: Thu, 25 Nov 2004 05:49:51 -0500 (EST)
• References: <200411230712.CAA29371@smc.vnet.net> <co1dp5\$sbm\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```DrBob <drbob at bigfoot.com> wrote:
[snip]
> In version 5.1 we have a different choice:
>
> (inputs)
> f[p_] = Piecewise[{{0, p == 0}, {(1/p) UnitStep[p - 1], True}}];
> Integrate[f@x, {x, -5, 5}]
> D[f@x, {x}]
> Integrate[%, x]
> Plot[f@x, {x, -5, 5}]
>
> (outputs)
> Log
> Piecewise[{{0, x < 1}, {-(1/x^2), x > 1}}, Indeterminate]
> Piecewise[{{0, x <= 1}}, -1 + 1/x]
> (and a plot)
>
> This time D followed by Integrate is evaluated, but WRONG.

I don't know why you think it's wrong. But in fact, it is not only right,
but also clever. Note that the original function is discontinuous, as is
its derivative. Then when we Integrate that, we need to get an
antiderivative of the derivative of f. But due to the discontinuity, an
antiderivative of the derivative of f need not be f itself. To confirm that
the result of Integrate is correct, just differentiate that result

Piecewise[{{0, x <= 1}}, -1 + 1/x]

Piecewise[{{0, x < 1}, {-(1/x^2), x > 1}}, Indeterminate]

just as desired?

The antiderivative is clever because Mathematica chose -1 as the constant
of integration in the part for x > 1, thereby nicely making the
antiderivative continuous.

David Cantrell

```

• References:
• UnitStep
• From: "Blimbaum, Jerry AA R22" <jerry.blimbaum@navy.mil>
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